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Mnenie [13.5K]
3 years ago
15

PLEASE PLEASE HELP!

Chemistry
1 answer:
valentinak56 [21]3 years ago
8 0

Answer: The number of grams of H_2 in 1620 mL is 1.44 g

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = 1620 ml = 1.62 L  (1L=1000ml)

n = number of moles = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =273K

n=\frac{PV}{RT}

n=\frac{1atm\times 16.2L}{0.0821Latm/K mol\times 273K}=0.72moles

Mass of hydrogen =moles\times {\text {Molar mass}}=0.72mol\times 2g/mol=1.44g

The number of grams of H_2 in 1620 mL is 1.44 g

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How many moles of carbon dioxide gas should be produced when 10.0 g of C2H6 are combusted at STP?
Finger [1]

Answer:

                    0.665 moles of CO₂

Explanation:

                     The balance chemical equation for the combustion of Ethane is as follow:

                            2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

Step 1: <u>Calculate moles of C₂H₆ as;</u>

                              Moles  =  Mass  /  M.Mass

Putting values,

                              Moles  =  10.0 g / 30.07 g/mol

                              Moles  =  0.3325 moles

Step 2: <u>Calculate Moles of CO₂ as;</u>

According to balance chemical equation,

                    2 moles of C₂H₆ produced  =  4 moles of CO₂

So,

             0.3325 moles of C₂H₆ will produce  =  X moles of CO₂

Solving for X,

                      X  =  0.3325 mol × 4 mol ÷ 2 mol

                      X = 0.665 moles of CO₂

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2. 

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